Simple pendulum question

1. A pendulum with length L is hung at the roof of a lift. What would the period of the pendulum be
when the lift goes up and down with constant acceleration a, respectively?
Explanation:
When the lift goes up with constant acceleration a, according the Newton’s second law, the force on
the pendulum will be F-mg=ma (Figure 1). F=mg+ma=m(g+a). So the new “acceleration” of the
pendulum will be F/m=g+a. It is the same as the lift going down, so the new acceleration will be
g-a.
F
a
G=mg
Figure 1
Solution:
Going up: F-mg=ma, F=mg+ma=m(g+a), a’= F/m=g+a
So, the period is T=2π√
Going down: F=mg-ma=m(g-a), a’= F/m=g-a
So, the period is T=2π√